Question

Mike started a savings account by depositing $9. Each month he deposits more money than the month before. At the end of 41 months, he has saved $9,204.50. How much more does he deposit each month?
@ranga

Answer 1

@ranga

Answer 2

@Disco619

Answer 3

Assume the money deposited is increasing in arithmetic progression and see if you can find the common difference. Use the formula for the sum of an arithmetic sequence with 41 terms and starting term 9 and common difference d.

Answer 4

Is interest accruing in this account or not?
I think there are too many unknowns.
Even without interest, how do you know whether to use an arithmetic series or a geometric series, or neither?

Answer 5

How much MORE does he deposit each month?

We are making the assumption they are not really expecting us to come up with a list of 40 differences in amounts. So we are making the assumption each month the deposit amount differs from the previous month by the same amount which suggests arithmetic series.

Since there is no mention of interest rate we have to assume this is a non-interest bearing savings account.

Answer 6

Ok, fair enough, but you can have a geometric series and say he deposits some number, say 1.45 or whatever it turns out to be, times more each time than the previous time. You could have a geometric series, and it would be relatively easy to describe although each new deposit would increase from the previous one by a fixed factor, and not a fixed difference.

Answer 7

In fact, I worked the problem out with both a geometric and an arithmetic series and got solutions for both.

Answer 8

Correct. This problem is too vague. If it is neither a AS or GS I would simply make the deposit $1 more each month and in the final month make a huge deposit to add up to $9,204.50.

Answer 9

@ranga I think you are correct.
I think the intended answer is an arithmetic series. There is a relatively simple number (with 3 decimal places) that will give it the exact sum they are looking for. Using a geometric series will take a number (with 6 decimal places ) that will give a sum that is approximately the number they are looking for.

Answer 10

The problem is that since the additional monthly amount (in dollars) has 3 decimal places, this could not be done because there is no way to deposit a fraction of one cent.

Answer 11

whats the formula? i use \[s _{n}=\frac{ n }{ 2 }\left[ 2a _{1} +(n-1)d \right]\]

Answer 12

@ranga

Answer 13

Yes that is the formula to use for the sum of an arithmetic series.

Answer 14

how do i plug in the numbers?

Answer 15

S_n = 9,204.50 ; n = 41 ; a_1 = 9 ; Find d.

Answer 16

ok wait i will solve it

Answer 17

ok so far I have \[9,204.50=41/2 [58(d)]\]

Answer 18

what do i do now?

Answer 19

It should be : 9204.5 = 41/2 * [ 18 + 40*d ]

Answer 20

i added the 18 + 40

Answer 21

Can't do that. 40 has a d attached to it. You can only add like terms.

Answer 22

oh so what do i do now

Answer 23

9204.5 = 41/2 * [ 18 + 40*d ]
9204.5 = 20.5 * [ 18 + 40*d ]
divide both sides by 20.5
9204.5/20.5 = [ 18 + 40*d ]
449 = 18 + 40d
subtract 18 from both sides:
449-18 = 40d
431 = 40d
divide both sides by 40:
431/40 = d
d = 10.775
round it to the nearest cent.
d = $10.78

How much more does he deposit each month? He deposits $10.78 more each month.

Answer 24

these are the options
$11.00

$11.50

$12.00

$12.50

Answer 25

THANK YOUUUU

Answer 26

Rounded to the nearest dollar (or even rounded to the nearest 50 cents), it is $11.00.

But you can take the answer $11 and go back to the original problem and add $11 each month more to the deposit and you will find the savings will go a few dollars past $9204.50.

That will be true of all the answer choices. The others will go way past.

You are welcome.